3

So Pokemon and McDonald's recently had a promotion where with each Happy Meal, there would be a pack of Pokemon cards included. Each pack contains a holo card. There are 25 holo cards in all. How many Happy Meals are you expected to buy? Now instead of 25 holo cards, let's say you'll be satisfied with collecting 15 unique ones, or 5 unique ones. Well in that case, how may Happy Meals do you need to buy? I created a Monte Carlo simulation here to solve that. The x-axis represents how many unique holos you want, and the y-axis represents the average number of Happy Meals you're expected to buy to obtain that many. Now, last night I was thinking, if we've already collected 24 unique holos, since we're looking for 1 last unique holo, we have a $\frac{1}{25}$ chance of opening it, or in other words, we're expected to open 25 packs before we find it. That would explain why in the graph in my simulation, the y-axis jumps from 70 to 95 when x goes from 24 to 25. That got me thinking, that's when we have 1 card left, and there's 25 cards in all. If we have 0 cards, then the chances we open a unique holo is $\frac{25}{25}$, as there are 25 cards we haven't collected, and 25 cards in all. So then I did a sum:

$$\frac{25}{25} + \frac{25}{24} + \frac{25}{23} ... + \frac{25}{1}$$

Essentially $$\sum_{n=0}^{24} \frac{25}{25-n}$$

Amazingly, this sum is 95.3, roughly the number obtained in my Monte Carlo Simulation. Similarly, if we changed the upper limit, we get what is roughly each point on from my simulation. I'm not really that good at probability or combinatorics. What I'm confused about, is why does this work? If we have 23 unique holos, why is the expected number of Happy Meals to open the next unique holo $\frac{25}{2}$? Or if we have 1 unique holo, why is the expected number of Happy Meals to open the next unique holo $\frac{25}{24}$? Thank you all so much for reading this far, I can't believe a stupid Pokemon promotion is getting me to think about math this much.

RobPratt
  • 45,619
therealnemis
  • 31
  • 2
    This problem is better known as the 'coupon collector's problem'. Basically, the probability of opening the $k+1$-st new 'coupon' out of a collection of $n$ unique coupons given you already have $k$ is geometrically distributed as $p_{k+1} = \frac{n-k}{n}$, so the expected value of this event is $\frac{1}{p_{k+1}}$, and by linearity of expectation we get your result. – lc2r43 Apr 22 '21 at 03:44
  • Hahaha wow, I'm reading it on Wikipedia right now. Thank you so much – therealnemis Apr 22 '21 at 03:48
  • 1
    The Wikipedia article has lots of useful info, but it's great that you've already tried thinking about it, and attacking it with a simulation. The mathematics for collecting a single set isn't too hard to understand, but it gets a bit trickier when you want to collect multiple sets. – PM 2Ring Apr 22 '21 at 03:52
  • I sort of get an understanding of the problem as well as a bit of the math that goes behind it. Now I'm wondering, why is the expected value the reciprocal of the geometric distribution? That's something I'm still a little lost on – therealnemis Apr 22 '21 at 03:56
  • Back in the real world, the cards are probably not equally likely so you have to keep going much longer. – badjohn Apr 22 '21 at 07:20

0 Answers0